# If the mol ratio of nitrogen and oxygen is #4:1#, what is the ratio of their solubilities in terms of mol fractions? Take #k_H = 3.3 xx 10^7 "torr"# for #"N"_2(g)# and #6.6 xx 10^7 "torr"# for #"O"_2(g)#.

##### 1 Answer

You should consider whether the question wants

There are two things you should be thinking about here:

- What is Henry's law in terms of mol fractions?
- What does knowing the ratio of
#"N"_2:"O"_2# in air tell you about their partial pressures?

Setting up the solution is usually the hard part. Imagine a beaker of water that has air dissolved in it, but there is, say, parafilm on it so that it is a closed container. Kind of like this, but with transparent water:

What you are looking for is the ratio of the solubilities on the mol fraction scale. It may or may not be

**Henry's law** in terms of mol fractions for the *solute* in particularly-dilute solutions is:

#bb(P_j = chi_(j(v))k_(H,j))# where:

#P_j# is thepartial pressureof solute#j# above the solution.#chi_(j(v))# is themol fractionof solute#j# in the vapor phase (#v# ), that is,abovethe solution.#k_(H,j)# is theHenry's law constantof solute#j# , and#k_(H,j)# approaches the pure vapor pressure#P_j^"*"# as#chi_(j(l)) -> 1# (since the solute particles surround the solvent particles completely, and vaporize most easily).

Writing Henry's law for

#P_(N_2) = chi_(N_2(v))k_(H,N_2)#

#P_(O_2) = chi_(O_2(v))k_(H,O_2)#

But from Dalton's law of partial pressures for ideal gases, we have:

#P_(t ot) = P_(N_2) + P_(O_2)#

#= 0.78P_(t ot) + 0.21P_(t ot) + stackrel("Other")overbrace(0.01P_(t ot))#

Thus, we can rewrite Henry's law as:

#0.78P_(t ot) = chi_(N_2(v))k_(H,N_2)#

#0.21P_(t ot) = chi_(O_2(v))k_(H,O_2)#

Finding the ratio of their mol fraction solubilities, we then have:

#(0.78P_(t ot))/(0.21P_(t ot)) = (chi_(N_2(v))k_(H,N_2))/(chi_(O_2(v))k_(H,O_2))#

#=> (chi_(N_2(v)))/(chi_(O_2(v))) = 3.71(k_(H,O_2)/k_(H,N_2))#

#= 3.71((3.3 xx 10^7 "torr")/(6.6 xx 10^7 "torr"))#

#= 1.86#

That's using a mol ratio of

#color(blue)((chi_(N_2(v)))/(chi_(O_2(v)))) = 4((3.3 xx 10^7 "torr")/(6.6 xx 10^7 "torr"))#

#= color(blue)(2:1)# ,#color(blue)("N"_2:"O"_2)#

If the question wants the ratio of

**In the end, it is just the mol ratio in air, weighted by the reciprocal ratio of the Henry's law constants.**